A research chemist has two different salt water solutions. Solution 1 is 25% salt and solution 2 is 20% salt. She wants to have 1 liter of 22% solution. How much of each solution should she mix together to get the desired solution?
amount of solution #1 --- x L
amount of solution #2 ---- 1-x L
.25x+ .20(1-x) = .22(1)
solve for x
btw, this would not be considered Calculus.
To determine how much of each solution the research chemist should mix, we can use the concept of the "mixture problem" and set up a system of equations.
Let's assume x represents the amount of solution 1 (25% salt) needed, and y represents the amount of solution 2 (20% salt) needed to obtain a 1 liter mixture of 22% salt.
There are a few key observations we can make to create the equations for the system. First, the total volume of the mixture will be 1 liter:
x + y = 1 -- Equation 1
Second, the total amount of salt in the mixture can be determined by calculating the sum of the salt in each solution:
0.25x + 0.20y = 0.22(1) -- Equation 2
Now, we can solve this system of equations using substitution or elimination:
From Equation 1, we can express y in terms of x:
y = 1 - x
Substituting this value into Equation 2:
0.25x + 0.20(1 - x) = 0.22
Simplifying the equation:
0.25x + 0.20 - 0.20x = 0.22
Combining like terms:
0.05x + 0.20 = 0.22
Subtracting 0.20 from both sides:
0.05x = 0.22 - 0.20
0.05x = 0.02
Dividing both sides by 0.05:
x = 0.02 / 0.05
x = 0.4
Now, we can find the value of y by substituting the value of x into Equation 1:
y = 1 - x
y = 1 - 0.4
y = 0.6
So, the research chemist should mix 0.4 liters of solution 1 (25% salt) with 0.6 liters of solution 2 (20% salt) to obtain a 1 liter mixture with a salt concentration of 22%.